Use the closure properties of regular languages and a language $B$ known to be non-regular to prove that a language $A$ is not regular.
My understanding is that the closure properties only apply when both languages are regular. So, I'm not sure what such a proof would look like and I'm looking for an outline of what the proof would look like.
The class of regular languages is closed under intersection. Suppose that you can find a regular language $L$ such that $B=L\cap A$; if $A$ were regular, then $B$, being the intersection of two regular languages, would also be regular. However, we know that $B$ is not regular, so $A$ cannot be regular.
Other closure properties can be used in the same way.
